#1
Hey guys,

I'm retarded. Seriously, this stuff has gone right over my head.

The problem is :

A body is thrown upwards with an initial speed 'u' from a point 'P' which is 40m above the ground. After 6 seconds it is at a point 'Q' and its velocity 8m/s downwards. Find the value of 'u' and the height of 'Q' above thr ground.

The answers are meant to be 50.8m/s and 168.4m, can somebody explain this to me please ?

Thanks a lot UG
#3
Quote by Uncle Fonzie
The answer is 7


No, but then you get the divide by zero error and oh shi-
#5
Physics was such a long time ago...

Does it help that the acceleration on Earth is 9.8 m/s2? I had to do problems like that all the time and it helped to know the rate at which an object accelerates in free fall.
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#6
If you hang tight, I (or someone else) will figure this out.

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#7
velocity^2=2(change in position times acceleration)+initial velocity.
since the body is going straight up in the air, the acceleration is -9.8m/s^2 due to the earth's gravity.
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#8
By the way, are you neglecting friction due to air? Hope so.

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#9
Quote by SteveHouse
By the way, are you neglecting friction due to air? Hope so.


We are neglecting it, yeah.

Ugh, obviously I'm not mathematically minded at all ! Can't wrap my head around these things even slightly.

Thanks a lot for the help though guys
#10
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#11
Quote by ThroughtheHaze
velocity^2=2(change in position times acceleration)+initial velocity.
since the body is going straight up in the air, the acceleration is -9.8m/s^2 due to the earth's gravity.
voila.

That's my issue.... I can't remember the kinematics formulas. I give then.

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#12
The formulae are :

v=u+at
s=ut+1/2at
v^2=u^2+2as

(Taking:
u=initial velocity
v=final velocity
a=acceleration
s=displacement
t=time)

If that makes sense, they're probably taught differently everywhere though!
#13
Quote by ThroughtheHaze
velocity^2=2(change in position times acceleration)+initial velocity.
since the body is going straight up in the air, the acceleration is -9.8m/s^2 due to the earth's gravity.
voila.


the initial velocity should be squared also.

once you plug everything in it should be 64 = v0^2 + 2*9.8*(Q-40) but thats not enough to solve the equation. use the other equation: x1 = x0 +v0*t + 1/2*a*t^2 and plug everthing in for that and you get Q = 40 + v0*6 + 1/2*9.8*36.
use those 2 equations and solve for v0
#14
Quote by fizzzzzzzzzzzy
the initial velocity should be squared also.

once you plug everything in it should be 64 = v0^2 + 2*9.8*(Q-40) but thats not enough to solve the equation. use the other equation: x1 = x0 +v0*t + 1/2*a*t^2 and plug everthing in for that and you get Q = 40 + v0*6 + 1/2*9.8*36.
use those 2 equations and solve for v0


Cool, ok thanks a lot, but where did you get the two x's I bolded ? What do they represent ?

Thanks !
#15
x is another way to say position. (delta)x = change in distance, or total displacement as you called it somewhere. His x corresponds to your s.

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#16
Quote by Reapersoul
Hey guys,

I'm retarded. Seriously, this stuff has gone right over my head.

The problem is :

A body is thrown upwards with an initial speed 'u' from a point 'P' which is 40m above the ground. After 6 seconds it is at a point 'Q' and its velocity 8m/s downwards. Find the value of 'u' and the height of 'Q' above thr ground.

The answers are meant to be 50.8m/s and 168.4m, can somebody explain this to me please ?

Thanks a lot UG


What grade or level of high school/colledge are you in??? ThaT ALL SOUNDS CONFUZING WITH THE LETTERS AND THINGS...BUT I'M IN 8TH AND only doin algebra so i don't know any of this....(sorry for tha caps didn't relize i was doing it...)
#17
Quote by guitartaber93
What grade or level of high school/colledge are you in??? ThaT ALL SOUNDS CONFUZING WITH THE LETTERS AND THINGS...BUT I'M IN 8TH AND only doin algebra so i don't know any of this....(sorry for tha caps didn't relize i was doing it...)

Then try the BACKSPACE KEY. (That's "delete" on a Mac.) You are allowed to go back and erase what you wrote before you hit the Post button.

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#18
Make a list of all the variables, and put in the ones you know.

u = ?
v = -8 (negative, as I'm taking upwards to be the positive direction)
x = ?
t = 6
a = -9.8

Then pick an equation that will find one of the unknowns:

v = u + at
-8 = u - 58.8
u = 50.8 m/s

Then using that result, work out the other one:

v^2 = u^2 + 2ax
64 = 2580.64 -19.6x
19.6x = 2516.64
x = 128.4 m

x is the displacement, so the total distance from the ground, Q = 128.4 + 40 = 168.4m

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#20
First write down all the values of v, u, a, s and t

v=8
u=?
a=9.8
s=?
t=6

Then look at the formulae to see which of them you can use.

First part:

v=u+at
u=v-at
u=8-(9.8*6)
u= -50.8m/s (negative in this case means upwards)

Second part:

s=ut+0.5at^2
s=-50.8*6 + 9.8 * 36
s= -304.8 + 176.4
s= -128.4m (again, negative is up)

Add that t the starting height, 40m, and you get 168.4m

EDIT: Damn you mynamewontfit >=[
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#21
Oh righty, of course.

So

u=?
t=6
a=-9.8
v=-8
s=0

v=u+at
Sub in and you get 50.8, then take your answer and put it into

s=ut+1/2at^2
and it should work out.

OK, I think I have it. Thanks a lot guys!


Edit : LOL!! 3rd place !! ^^

Thanks loads guys, really helpful !